# College Math Teaching

## April 12, 2020

### A tidbit with respect to Laplace transforms and sin(x)/x

Filed under: complex variables, integrals, Laplace transform, media — collegemathteaching @ 9:01 pm

I’ve discovered the channel “blackpenredpen” and it is delightful.
It is a nice escape into mathematics that, while far from research level, is “fun” and beyond mere fluff.

And that got me to thinking about $\int^{\infty}_0 \frac{sin(x)}{x} dx$. Yes, this can be done by residues

But I’ll look at this with Laplace Transforms.

We know that $\mathcal{L}(sin(x)) = \int^{\infty}_0 e^{-st}sin(t)dt = \frac{1}{s^2+1}$
But note that the antiderivative of $e^{-st}$ with respect to $s$ is $-\frac{1}{t}e^{-st}$ That might not seem like much help, but then notice $\int^{\infty}_0 e^{-st} ds = \frac{-1}{t}e^{-st}|^{\infty}_0 = \frac{1}{t}$ (assuming $s > 0$

So why not: $\int^{\infty}_0 \int^{\infty}_0 e^{-st}sin(t)dt ds = \int^{\infty}_0 \frac{1}{s^2+1} ds =arctan(s)|^{\infty}_0 = \frac{\pi}{2}$
Now since the left hand side is just a double integral over the first quadrant (an infinite rectangle) the order of integration can be interchanged:
$\int^{\infty}_0 \int^{\infty}_0 e^{-st}sin(t)dt ds = \int^{\infty}_0 \int^{\infty}_0 e^{-st}sin(t)ds dt = \int^{\infty}_0 sin(t) \int^{\infty}_0 e^{-st}ds dt = \int^{\infty}_0 sin(t)\frac{1}{t} dt$

and that is equal to $\frac{\pi}{2}$.

Note: $\int_0^x\frac{sin(t)}{t} dt$ is sometimes called the $Si(x)$ function

\

## April 6, 2020

### How I am cutting corners in class

Filed under: complex variables, differential equations, Laplace transform — collegemathteaching @ 12:06 am

Ok, which is more difficult?

1. Solve $x'' + 6x' + 13x = sin(t), x(0) = x'(0) = 0$ using Laplace transforms or:

2. Given $Y = \frac{1}{s^4 + 6s^3 +10s^2 + 6s + 9}$ find the inverse Laplace transform.

Clearly, 2 is harder and in texts I’ve used, we had to do those prior to doing 1. But, in a way, you have to do 2 in order to do 1:

$(s^2 + 6s + 13)X(s) = \frac{1}{s^2+1} \rightarrow X(s) = \frac{1}{(s^2 + 6s +13)(s^2 + 1)}$

But this is already factored and students can be taught to “attempt to factor and if you can’t, complete the square” and this leads immediately to:

$X(s) = \frac{1}{(s^2 + 6s + 9 +4)(s^2+1)} = \frac{As + B}{(s+3)^2+4} + \frac{Cs + D}{s^2 +1}$ which can be resolved by partial fractions.

In our “one less week plus online” I will do much more of 1 than 2.

Of course, there is still some work to do; we still have to solve $(As+B)(s^2+1) +(Cs+D)((s+3)^2 +4) =1$

I will teach the “eliminate the term method by using complex numbers:

Let $s = i$ to get

$(Ci+D)(12+6i) = 12D-6C +(12C + 6D)i = 1 \rightarrow D = -2C \rightarrow -30C = 1 \rightarrow C = -\frac{1}{30}, D = \frac{1}{15}$
Let $s = -3+2i$

$\rightarrow s^2+1 = 6-12i \rightarrow (-3A+B +2iA)(6-12i)$

$= -18A+6B +24A +36iA+12Ai-12Bi = -6A+6B +(48A-12B)i = 1$

$\rightarrow B=4A, 6A+6B= 1\rightarrow 30A=1$

So we have $A = \frac{1}{30}, B = \frac{2}{15}, C = -\frac{1}{30}, D = \frac{1}{15}$

The particular solution part pulls back to $-\frac{1}{30} cos(t) + \frac{1}{15} sin(t)$

There is a work to do for the other part:

To get the $s+3$ shift we have to add and subtract 3; this leads to:

$\frac{A(s+3) + B-3A}{(s+3)^2+4} =\frac{A(s+3) }{(s+3)^2+4} + \frac{ B-3A}{(s+3)^2+4}$

$\frac{1}{30}\frac{(s+3) }{(s+3)^2+4} + \frac{1}{30}\frac{1}{2}\frac{2}{(s+3)^2+4}$ (adjusting the second term for the $4 = 2^2$
And this part pulls back to $\frac{1}{30}e^{-3t}cos(2t) +\frac{1}{60} e^{-3t}sin(2t)$

Yeah, I know; if you are reading this, you already know this stuff, but I think using i helps speed things up a bit.

And yes, you could have just used the convolution integral and have been done with it, though one would have had to have used
$\frac{1}{2}e^{-3t}sin(2t) * sin(t) =\int^t_0\frac{1}{2}e^{-3u}sin(2u)sin(2t-2u)du$ and been done with it. (you remembered the 1/2, didn’t you? )